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Problem 2.5: Find the electric field a distance $z$ above the center of a circular loop of radius $r$ which carries a uniform line charge $\lambda$.

This problem is in refereced here (with solution):

But it is solved using a trick. The trick is: okay we know there is symmetry so let us just take the magnitude and not use the vectors. The problem I have is I am unable to solve it without using the trick and I would appreciate someones help in doing the problem without the shortcut, using the vectors. When I do it the r compontent doesn't cancel.

Using polar coordinates here is the setup:

$$dE = \frac{1}{4 \pi \epsilon_0} \frac{\lambda}{\rho^2} ~ \hat{\rho} ~ r d \theta$$

$$\hat{\rho} = \frac{z \hat{k} + r \hat{r}}{\sqrt{z^2 + r^2}} $$

$$ E = \frac{\lambda}{4 \pi \epsilon_0} \int_0^{2 \pi} \frac{r z \hat{z} + r^2 \hat{r} }{(z^2 + r^2)^{3/2}} ~ d \theta$$

$$ E = \frac{\lambda}{4 \pi \epsilon_0} \int_0^{2 \pi} \frac{r z \hat{z} }{(z^2 + r^2)^{3/2}} ~ d \theta + \frac{\lambda}{4 \pi \epsilon_0} \int_0^{2 \pi} \frac{r^2 \hat{r} }{(z^2 + r^2)^{3/2}} ~ d \theta $$

If I now evaluate the $z$ component I get the correct answer. When I evaluate the $r$ component, shown below the $r$ doesn't not cancel the way it should. why not?

$$\frac{\lambda}{4 \pi \epsilon_0} \frac{r^2 2 \pi }{(z^2 + r^2)^{3/2}} - \frac{\lambda}{4 \pi \epsilon_0} \frac{r^2 0 }{(z^2 + r^2)^{3/2}} $$

$$ = \frac{\lambda}{4 \pi \epsilon_0} \frac{r^2 2 \pi }{(z^2 + r^2)^{3/2}} \hat{r}$$

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closed as too localized by dmckee May 31 '13 at 0:37

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note that $\hat{r}$ is not a constant vector! – Ali May 28 '13 at 17:05
Duplicate :… – ABC May 29 '13 at 3:11
up vote 2 down vote accepted

Recall that $\hat r$ depends on $\theta$ in the following way: $$ \hat r = \cos\theta \,\hat x +\sin\theta \,\hat y $$ It follows that $$ \int_0^ {2\pi }\hat r \,d\theta = \hat x \int_0^ {2\pi}\cos\theta \, d\theta + \hat y \int_0^{2\pi}\sin\theta \,d\theta = 0+0 $$

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